# Conjugacy of $L_X$ operators

Assume that $X$ is a vector field on a $n$ dimensional manifold $M$.Let $0\leq i,j \leq n$.

1.Is there a linear isomorphism $T:\Omega^i(M) \to \Omega^j(M)$ with $L_X T=T L_X$?

2.Is there a linear isomorphism $T:\Omega^i(M)/Z^i(M) \to \Omega^j(M)/Z^j(M)$ with $TL_X=L_XT$?

By $Z^i(M)$ we mean the space of closed differential $i$_form.

For a motivation please see the following post:

Fredholm index vs. Limit cycle theory

• The spaces you consider need not be isomorphic, so in general, this answer is no. – Matthias Ludewig May 21 '17 at 15:24
• @MatthiasLudewig What is a precise example of this non isomorphicity? – Ali Taghavi May 21 '17 at 16:34

For appropriate choices of $i$, $j$ (e.g. $i+j \neq n$), $\Omega^i(M)$ and $\Omega^j(M)$ have different ranks as $C^\infty(M)$ module, so at least they cannot be isomorphic as modules. Maybe they could be isomorphic as topological vector spaces, but I don't see why, and in any case, this would not be very natural.
If you choose $i=j$, then of course $T = \mathrm{id}$ works.
More generally, setting $T= \psi_t^*$ for any $t \in \mathbb{R}$ does the job, where $\psi_t$ is the flow of $X$, and $\psi_t^*$ denotes pullback via the flow.
• @MittihasLudewig (+1) and thank you for your answers to my questions. I think that it is natural to compare $\Omega^{i}(M)$ and $\Omega^{j}(M)$ as two vector spaces not two modules, since $L_{X}$ is not a $C^{\infty}(M))$-linear operator. – Ali Taghavi May 22 '17 at 11:08